Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468

First let’s find the smallest number which is exactly divisible by both 520 and 468. That is simply just the LCM of the two numbers. By prime factorisation, we get 520 = 2 3 × 5 × 13 468 = 2 2 × 3 2 × 13 ∴ LCM (520, 468) = 2 3 × 3 2 × 5 × 13 = 4680 Hence, 4680 is the smallest number which is exactly divisible by both 520 and 468 i.e. we will get a remainder of 0 in each case. But, we need to find the smallest number which when increased by 17 is exactly divided by 520 and 468. So that is found by, 4680 – 17 = 4663 ∴ 4663 should be the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is obtained by subtracting 17 from the LCM of 468 and 520. Prime factorization of 468 and 520 is: 468 = 2 2× 3 2× 13 520 = 2 3× 5 × 13 LCM = product of greatest power of each prime factor involved in the numbers = 2 2× 3 2× 5 × 13 = 4680 The required number is 4680 – 17 = 4663. Hence, the smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663.

In mathematics, we often encounter problems that require us to find the smallest or largest number that satisfies certain conditions. One such problem involves finding the smallest number that, when increased by a certain value, becomes divisible by two given numbers. In this article, we will explore how to find the smallest number which when increased by 17 is exactly divisible by both 520 and 468, we can use the following trick: • Find the LCM of 520 and 468. The LCM of 520 and 468 is 4680. • Solve (x + 17) = k * LCM , where x is the smallest number • Replace values of k above and check if divisible by 468 and 520 both Step 1 : Find the LCM of 520 and 468 Prime Factorization of 468 is: 2 x 2 x 3 x 3 x 13=>2 2 x 3 2 x 13 1 Prime Factorization of 520 is: 2 x 2 x 2 x 5 x 13=>2 3 x 5 1 x 13 1 For each prime factor, find where it occurs most often as a factor and write it that many times in a new list. The new superset list is 2, 2, 2, 3, 3, 5, 13 Multiply these factors together to find the LCM. LCM = 2 x 2 x 2 x 3 x 3 x 5 x 13 = 4680 In exponential form: LCM = 2 3 x 3 2 x 5 1 x 13 1 = 4680 LCM = 4680 Therefore, LCM(468, 520) = 4680 Step 2 : Solve for (x + 17) = k * LCM Let the smallest number be x. Then, we have the equation: (x + 17) = k * LCM where k is an integer. Substituting the LCM, we get: x + 17 = k *4680 For k=1 x+17 = 4680 x= 4663 , which is divisible by 520 and 468 Finding the smallest number which when increased by 17 is exactly divisible by both 520 and 468 is a...

Hint: In the given question, we are required to find the smallest number which when increased by $ 17 $ is exactly divisible by both $ 520 $ and $ 468 $ . So, we first try to find the smallest number divisible by both $ 520 $ and $ 468 $ by taking the least common multiple of both the numbers. Then, we find our required answer by subtracting $ 17 $ from the LCM of $ 520 $ and $ 468 $ as the number has to be increased by $ 17 $ to be the LCM of the give numbers. Complete step-by-step answer: So, we first find the LCM of the numbers $ 520 $ and $ 468 $ . To find the least common multiple of $ 520 $ and $ 468 $ , first we find out the prime factors of both the numbers. Prime factors of \[520\] $ = 2 \times 2 \times 2 \times 5 \times 13 $ $ = \times 5 \times 13\] $ = 4680 $ Hence, the least common multiple of $ 520 $ and $ 468 $ is $ 4680 $ . Now, we know that we have to find the smallest number which when increased by $ 17 $ is exactly divisible by both $ 520 $ and $ 468 $ . So, the smallest number which is divisible by both $ 520 $ and $ 468 $ is $ 4680 $ as the LCM of $ 520 $ and $ 468 $ . So, to get our required answer, we have to subtract $ 17 $ from the LCM of $ 520 $ and $ 468 $ , that we found as $ 4680 $ . So, the required answer is $ 4680 - 17 = 4663 $ . So, the correct answer is “4663”. Note: Least common multiple (LCM) has wide ranging applications in real world as well as in mathematical questions. Knowledge of the least common multiple is also used in addition a...

The given numbers are 520 and 468. The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the LCM of 520 and 468. Prime factorisation of 520 = 2 × 2 × 2 × 5 × 13 Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13 LCM of 520 and 468 = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680. Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 – 17 = 4663.

The greatest four digit number = 9 9 9 9 we can get the required number by following steps: step 1 : Divide 9999 by LCM of 1 5 , 2 4 and 3 6 Step 2 : Subtract result of step 1 from 9 9 9 9 . Prime factorization : 1 5 = 3 × 5 2 4 = 2 3 × 3 3 6 = 2 2 × 3 2 LCM ( 1 5 , 2 4 , 3 6 ) = 2 3 × 3 2 × 5 = 3 6 0 Step 1 : Dividing 9 9 9 9 by 3 6 0 , we get 2 7 9 as remainder. Step 2 : subtract 2 7 9 from 9 9 9 9 9 9 9 9 − 2 7 9 = 9 7 2 0 the greatest number of four digit which is exactly divisible by 1 5 , 2 4 and 3 6 is 9 7 2 0 .